When each of x, y, and z minus the product of the three gives a square

 
 
Find rational numbers   x, \; y, \; z   such that :

x \; - \; x \cdot y \cdot z \; = \; A^2
y \; - \; x \cdot y \cdot z \; = \; B^2
z \; - \; x \cdot y \cdot z \; = \; C^2

 
 

Here’s the first solution:

(1/9),   (441/625),   (1296/1369)   are the first three squares that work.

x = (1/3)^2 = (1/9) ….. y = (21/25)^2 = (441/625) ….. z = (36/37)^2 = (1296/1369)

 

x \cdot y \cdot z \; = \; (1/9) \cdot (441/625) \cdot (1296/1369) \; = \; 63504/855625 \; = \; (252/925)^2

(1/9) \; - \; (252/925)^2 \; = \; 284089/7700625 \; = \; (533/2775)^2

(441/625) \; - \; (252/925)^2 \; = \; 21609/34225 \; = \; (147/185)^2

(1296/1369) \; - \; (252/925)^2 \; = \; 746496/855625 \; = \; (864/925)^2

 
 
Find other solutions
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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